Question

Solve the following differential equation:

${\displaystyle \left(\text{x + y}\right)\frac{\text{dy}}{\text{dx}}=1}$

Answer 1

${\displaystyle \left(\text{x + y}\right)\frac{\text{dy}}{\text{dx}}=1}$

∴ ${\displaystyle \frac{\text{dx}}{\text{dy}}=\text{x + y}}$

∴ ${\displaystyle \frac{\text{dx}}{\text{dy}}-\text{x}=\text{y}}$

∴ ${\displaystyle \frac{\text{dx}}{\text{dy}}+(-1)\text{x}=\text{y}}$      ....(1)

This is the linear differential equation of the form

${\displaystyle \frac{\text{dx}}{\text{dy}}+\text{P}\cdot \text{x}=\text{Q},}$ where P = - 1 and Q = y

∴ I.F. = ${\displaystyle {\text{e}}^{\int \text{P dy}}={\text{e}}^{\int -1\text{dy}}={\text{e}}^{-\text{y}}}$

∴ the solution of (1) is given by

x.(I.F.) = ${\displaystyle \int \text{Q}\cdot \left(\text{I.F.}\right)\text{dy}+\text{c}}$

∴ ${\displaystyle \text{x}\cdot {\text{e}}^{-\text{y}}=\int \text{y}\cdot {\text{e}}^{-\text{y}}\text{dy}+\text{c}}$

∴ ${\displaystyle {\text{e}}^{-\text{y}}\cdot \text{x}=\text{y}\int {\text{e}}^{-\text{y}}\text{dy}-\int [\frac{\text{d}}{\text{dx}}\left(\text{y}\right)\int {\text{e}}^{-\text{y}}\text{dy}]\text{dy}+\text{c}}$

${\displaystyle =\text{y}\cdot \frac{{\text{e}}^{-\text{y}}}{-1}-\int 1\cdot \frac{{\text{e}}^{-\text{y}}}{-1}\text{dy}+\text{c}}$

${\displaystyle =-{\text{ye}}^{-\text{y}}+\int {\text{e}}^{-\text{y}}\text{dy}+\text{c}}$

∴ ${\displaystyle {\text{e}}^{-\text{y}}\cdot \text{x}=-{\text{ye}}^{-\text{y}}+\frac{{\text{e}}^{-\text{y}}}{-1}+\text{c}}$

∴ ${\displaystyle {\text{e}}^{-\text{y}}\cdot \text{x}+{\text{ye}}^{-\text{y}}+{\text{e}}^{-\text{y}}=\text{c}}$

∴ ${\displaystyle {\text{e}}^{\text{-y}}\left(\text{x + y + 1}\right)=\text{c}}$

∴ x + y + 1 = ce^y 

This is the general solution.